Question

Two beads, each with charge q and mass m, are on a horizontal, frictionless, non-conducting, circular hoop of radius R. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by [ \(\epsilon_0 \)is the permittivity of free space.]

• A. \(\frac{q^2}{4\pi\epsilon_0R^3m}\)
• B. \(\frac{q^2}{32\pi\epsilon_0R^3m}\)
• C. \(\frac{q^2}{8\pi\epsilon_0R^3m}\)
• D. \(\frac{q^2}{16\pi\epsilon_0R^3m}\)

Answer 1

The Correct Option is B

The equilibrium position of the oscillating bead is diametrically opposite to the fixed bead, because that is where the electrostatic force is minimized (and potential energy is minimized). Let the angular displacement from this equilibrium position be \(\theta\). The distance between the beads is then given by using the law of cosines: \[r = \sqrt{R^2 + R^2 - 2R^2 \cos(\pi - \theta)} = \sqrt{2R^2(1 - \cos(\pi - \theta))} = \sqrt{2R^2(1 + \cos(\theta))}\] For small \(\theta\), we can approximate \(\cos(\theta) \approx 1 - \frac{\theta^2}{2}\). Therefore, \[r \approx \sqrt{2R^2 (1 + 1 - \frac{\theta^2}{2})} = \sqrt{4R^2 - R^2\theta^2} \approx 2R \sqrt{1 - \frac{\theta^2}{4}} \approx 2R(1 - \frac{\theta^2}{8})\]

The electrostatic potential energy is then: \[U = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r} \approx \frac{1}{4\pi\epsilon_0} \frac{q^2}{2R(1 - \frac{\theta^2}{8})} \approx \frac{q^2}{8\pi\epsilon_0 R} (1 + \frac{\theta^2}{8})\] The potential energy is approximately \[U = \frac{q^2}{8\pi\epsilon_0 R} + \frac{q^2 \theta^2}{64\pi \epsilon_0 R}\]

The kinetic energy of the oscillating bead is given by \[T = \frac{1}{2} m (R\dot{\theta})^2 = \frac{1}{2} m R^2 \dot{\theta}^2\]

The Lagrangian is \(L = T - U = \frac{1}{2} m R^2 \dot{\theta}^2 - \frac{q^2}{8\pi\epsilon_0 R} - \frac{q^2 \theta^2}{64\pi \epsilon_0 R}\).

The equation of motion can be derived using Euler-Lagrange equation, \(\frac{d}{dt}(\frac{\partial L}{\partial \dot{\theta}}) - \frac{\partial L}{\partial \theta} = 0\),

\[\frac{d}{dt}(m R^2 \dot{\theta}) - (-\frac{q^2 \theta}{32 \pi \epsilon_0 R}) = 0\] \[m R^2 \ddot{\theta} + \frac{q^2 \theta}{32 \pi \epsilon_0 R} = 0\] \[\ddot{\theta} + \frac{q^2}{32 \pi \epsilon_0 m R^3} \theta = 0\] Therefore, the angular frequency squared is

\[\omega^2 = \frac{q^2}{32 \pi \epsilon_0 R^3 m}\] Answer: (B) \(\frac{q^2}{32\pi\epsilon_0 R^3 m}\)

Answer 2

To solve the problem, we need to find the square of the angular frequency \(\omega^2\) of small oscillations of one charged bead on a frictionless circular hoop, with the other bead fixed and both having charge \(q\) and mass \(m\).

1. System Setup and Forces:
Two beads of charge \(q\) are on a circular hoop of radius \(R\). One bead is fixed, and the other can oscillate about its equilibrium position along the hoop. The beads repel each other due to electrostatic force.
The potential energy \(U(\theta)\) of the system depends on the electrostatic repulsion between the charges, where \(\theta\) is the angular displacement of the moving bead from the equilibrium.

2. Electrostatic Potential Energy:
The distance between the two charges when the moving bead is at angle \(\theta\) (measured from the fixed bead) on the hoop is:

\[ d = 2R \sin\left(\frac{\theta}{2}\right) \] The electrostatic potential energy between two charges is:

\[ U(\theta) = \frac{1}{4\pi\epsilon_0} \frac{q^2}{d} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{2R \sin(\theta/2)} = \frac{q^2}{8 \pi \epsilon_0 R \sin(\theta/2)} \]

3. Equilibrium Position:
The equilibrium position corresponds to the minimum potential energy. Since the beads repel each other, the moving bead will be opposite the fixed bead on the hoop, i.e., \(\theta = \pi\).

4. Small Oscillations Approximation:
For small oscillations about \(\theta = \pi\), let
\[ \theta = \pi + x, \quad |x| \ll 1 \] Using the small angle approximation for \(\sin\left(\frac{\theta}{2}\right)\):
\[ \sin\left(\frac{\pi + x}{2}\right) = \sin\left(\frac{\pi}{2} + \frac{x}{2}\right) = \cos\left(\frac{x}{2}\right) \approx 1 - \frac{x^2}{8} \] Thus,

\[ U(x) \approx \frac{q^2}{8 \pi \epsilon_0 R} \frac{1}{1 - \frac{x^2}{8}} \approx \frac{q^2}{8 \pi \epsilon_0 R} \left(1 + \frac{x^2}{8}\right) \]

5. Potential Energy Change and Effective Spring Constant:
Ignoring the constant term, the potential energy variation near equilibrium is:

\[ U(x) \approx U_0 + \frac{q^2}{8 \pi \epsilon_0 R} \times \frac{x^2}{8} = U_0 + \frac{q^2}{64 \pi \epsilon_0 R} x^2 \] Since \(x\) is angular displacement, the effective potential energy corresponds to a harmonic oscillator with:

\[ k_{\text{eff}} = \frac{d^2 U}{d x^2} = \frac{q^2}{32 \pi \epsilon_0 R} \] Note: The factor 2 in the second derivative comes from the coefficient of \(x^2\) term (coefficient is \(\frac{q^2}{64 \pi \epsilon_0 R}\)), so second derivative is twice this:
\[ k_{\text{eff}} = 2 \times \frac{q^2}{64 \pi \epsilon_0 R} = \frac{q^2}{32 \pi \epsilon_0 R} \]

6. Equation of Motion and Angular Frequency:
The moment of inertia for the bead of mass \(m\) moving on a circle of radius \(R\) is:

\[ I = m R^2 \] The angular frequency \(\omega\) of small oscillations is given by:

\[ \omega^2 = \frac{k_{\text{eff}}}{I} = \frac{q^2}{32 \pi \epsilon_0 R} \times \frac{1}{m R^2} = \frac{q^2}{32 \pi \epsilon_0 R^3 m} \]

Final Answer:
The square of the angular frequency of the small oscillations is \(\boxed{\frac{q^2}{32 \pi \epsilon_0 R^3 m}}\).